evaluating the economic efficiency of transportation ...srg/book/files/pdf/8....

1K u m a r e s C. S I n h a a n d S a m U E L L a b I

Evaluating the Economic Efficiency of Transportation Interventions

Costs $

Benefits $

Transportation Decision-making – Principles of Project Evaluation and Programming

Contents of this Presentation

Cash Flow Illustrations

The 5 Variables in any Cash Flow Diagram

Equivalence Equations (Interest Formula)

Criteria for Economic Efficiency Evaluation

Special Cases of Equivalence Equation

Steps for Economic Efficiency Analysis

What is Life-cycle Cost Analysis (LCCA)?2

Economic Efficiency

Why Carry out Economic Efficiency Analysis?

Economic efficiency is a key criterion for evaluation of most engineering systems

Useful for evaluating systems on basis of 1 or more PM that are monetizeable

Is a type of multiple-objective decision analysis

Reflects that fact that the value of money changes over time (Example: $1,000 in today is not the same as $1,000 in 1993, and is not the same as $1,000 in 2013)

Concept of cash flow over time

3

Economic Efficiency

The Concept of Cash Flows

Is a representation of money inflows and outflows assoc. with an engineering system

Table or Diagram can be used

Table: +ve money coming in (revenues, savings)-ve money going out (costs)

Diagram: downward arrows money coming in (revenues, savings)upward arrows money going out (costs)

Time intervals of money flows: daily, weekly, monthly, yearly, etc.

4

Economic Efficiency

Cash Flow Diagrams for Transportation Projects

For CE systems, intervals are mostly Yearly

Amounts received or incurred during year are assumed to have happened …

… at end of year, or … at beginning of year(choose only 1 of above conventions, and stick to it)

5

Economic Efficiency

Example 1:

Construction of a Freeway BypassConsider the following annual cash flows for a proposed freeway bypass.

Construction will be completed by December 2009 and the estimated annual benefits of travel time, safety, and VOC are provided below.

Year Construction Cost ($)

(Chapter 3)

Demand

(Chapter 4)

Travel Benefits ($)

(Demand and

Chapter 5)

Safety Time Benefits ($)

(Demand and

Chapter 6)

VOC Benefits ($)

(Demand and

Chapter 7) 2009 $35 M 2010 320,000 $5.76 M $1.33 M $0.60 M 2011 345,000 $6.12 M $1.43 M $0.65 M 2012 365,000 $6.57 M $1.51 M $0.69 M 2013 380,000 $6.84 M $1.58 M $0.72 M 2014 405,000 $7.29 M $1.68 M $0.76 M 2015 418,000 $7.52 M $1.73 M $0.79 M 2016 438,000 $7.88 M $1.82 M $0.83 M 2017 460,000 $8.28 M $1.91 M $0.87 M 2018 490,000 $8.82 M $2.03 M $0.92 M 2019 525,000 $9.45 M $2.18 M $0.99 M

Travel Time Benefits ($)

Safety Benefits ($)

6

Economic Efficiency

Cash Flow Diagram for Freeway Bypass Construction

35M

2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019

Construction Cost

7

Economic Efficiency


6.57M6.12M35M 5.76M 6.84M 7.29M 7.52M 7.88M 8.28M 8.82M 9.45M

2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019

Travel Time Benefits

8

Economic Efficiency


2.03M

6.57M6.12M35M 5.76M 6.84M 7.29M 7.52M 7.88M 8.28M 8.82M 9.45M

1.33M 1.43M 1.51M 1.58M 1.68M 1.73M 2.18M1.91M1.82M

2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019

Safety Benefits

9

Economic Efficiency


2.03M

6.57M6.12M35M 5.76M 6.84M 7.29M 7.52M 7.88M 8.28M 8.82M 9.45M

1.33M 1.43M 1.51M 1.58M 1.68M 1.73M 2.18M1.91M1.82M

0.60M 0.65M 0.99M0.92M0.69M 0.72M 0.76M 0.79M 0.83M 0.87M

2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019

VOC Benefits

10

Economic Efficiency

Example 2:

Post-implementation EvaluationReconstructed I-975 Bridge and Tollway

Initial Contract Cost (Year 1997)- $8 milToll receipts (1998-2005)- $2 mil per year Major Maintenance (Year 2000) - $3 milMinor Maintenance (Year 2004) - $1 mil

Benefits of Travel Time, Safety, VOCs: $4 mil/year (1997-2005) Operating costs (1997-2005)- $1.5 mil per year

Draw the Cash Flow Diagram for this example.11

Economic Efficiency

2M 2M2M 2M 2M 2M 2M 2M8M

3M 1M

4M

2003

Initial Contract Cost (Year 1997)- $8 milToll receipts (1998-2005)- $2 mil per year Major Maintenance (Year 2000) - $3 milMinor Maintenance (Year 2004) - $1 milBenefits of Travel Time, Safety, VOCs: $4 mil/year (1997-2005) Operating costs (1997-2005)- $1.5 mil per year

4M 4M 4M 4M 4M 4M 4M

1.5M 1.5M 1.5M 1.5M 1.5M 1.5M 1.5M 1.5M

1997 1998 1999 2000 2001 2002 2004 2005

12

Economic Efficiency

The 5 Variables in any Cash Flow Diagram

P - Present AmountF - Future AmountA - Annual Amounti - Interest RateN - Analysis Period

Typically, we’re given 4 of the above variables and we will need to calculate the 5th, using equivalence equations.

13

Economic Efficiency

Discussion of the 5 Variables

1. Present Amount, PAmount received or incurred in Year 0. Typically a large

amount, e.g., initial construction cost, initial money collected for car loan, etc.

2. Future Amount, FA future amount that is equivalent to a given present

amount (received or incurred).Also called “discounted amount”. May be incurred (or paid) at end of analysis period or

anytime within the analysis period.

14

Economic Efficiency

3. Annual Amounts, A

Amount received or incurred every period

Examples are operating costs and maintenance costs

Typically are “uniform amounts” but not always uniform. For example, O&M costs typically increases with system age or as demand increases

15

Economic Efficiency

4. Analysis Period, N

Is the total time over which we are carrying out economic evaluation for a system.

Synonyms: Payment Period, Planning Horizon, Planning Period.

Typically, expressed in years for CE systems.

Length depends on system type and expected life of the system.

May be the overall life of a system or only a part thereof

Examples: 35 years for highway pavements, 50 years for RC bridges, 70 years for steel bridges, 100 years for some dams. 16

Economic Efficiency

17

5. Interest Rate, i

Represents the extent to which the value of money changes over time.

Opportunity cost: Money used by a transportation agency for the project could’ve been used elsewhere to generate profit or used elsewhere to generate interest

Thus, by implementing the project, the agency is foregoing some benefit elsewhere (i is a measure of that benefit)

National interest rate is determined by the Central Bank (or Federal Reserve Board). Individual lenders also have their own rates, but are generally pegged to the federal rate.

Simple interest vs. compound interest

Economic Efficiency

Simple Interest vs. Compound Interest

1. Simple InterestLet’s say in Dec 2007 you borrow $1000 with 5% simple interest

Year Principal Interest Total Amount Owed at the End of The Year (=Principal + Interest)

Dec 2007 $1000 $1000

Dec 2008 $50 $1050

Dec 2009 $50 $1100

Dec 2010 $50 $1150

Dec 2011 $50 $120018

Economic Efficiency

2. Compound InterestLet’s say in Dec 2007 you borrow $1000 with 5% compound interest

Year Principal Interest Total Amount Owed at the End of The Year (=Principal + Interest)

Dec 2007 $1000 - $1000

Dec 2008 $1000 $50 $1050

Dec 2009 $1050 $52.5 $1102.5

Dec 2010 $1102.5 $55.13 $1157.63

Dec 2011 $1157.63 $57.88 $1215.5119

Economic Efficiency

Simple Interest vs. Compound Interest

Year Simple Interest

Compound Interest

Dec 2007 $1000 $1000

Dec 2008 $1050 $1050

Dec 2009 $1100 $1102.5

Dec 2010 $1150 $1157.63

Dec 2011 $1200 $1215.51

Total Amount Owed at the End of The Year (=Principal + Interest)

20

Economic Efficiency

Compound interest (not simple interest) is often used.

Compound interest may be fixed or variable

Fixed (constant) interest: same interest rate during the year

same interest rate in each of several years

Variable interest rate: interest rate changes during the year.

Change may be as follows:

- compounding a finite number of times during each year

- compounding an infinite number of times each year21

Economic Efficiency

The 5 Variables …

P - Present AmountF - Future AmountA - Annual Amounti - Interest RateN - Analysis Period

Given any 4 Find value of 5th variable

Use “Equivalence equations”(also called “Interest formula”)

22

Economic Efficiency

Equivalence Equations …Used for calculating any 1 of the 5 Cash Flow Variables, given the other 4.

Two main categories of Equivalence Equations, mostly depending on the nature of the interest rate (how it changes).

Equivalence Equations

23

Type 1Interest rate is:

- Constant, OR

- Variable but compounded a finite number of times within the year

Type 2Interest rate is:

- Variable but compounded an infinite number of times within the year

Economic Efficiency

Equivalence Equations Type I

24

Economic Efficiency

Description: Finding the future amount (F) to be yielded by an initial amount (P) at the end of a given period.

Example: Current car price is $20,000. Jim takes the car now (Dec 2007) and agrees with the dealer to pay nothing until Dec 2012 when he pays everything at a go.

What price will he pay in 2012? Assume 5% interest rate.

Case 1

25

Economic Efficiency

26

Case 1 (cont’d)Problem Definition:

This is a Single Payment Compound Amount Factor (SPCAF) problem

Cash Flow Diagram:

Computational Formula:2007 2012

P F = ?

])1[( niPF +×=SPCAF

Economic Efficiency

Solution to Case 1 Example:

Question: Current car price is $20,000. Jim takes the car now (Dec 2007) and agrees with the dealer to pay nothing until Dec 2012 when he pays everything at a go.

What price will he pay in 2012? Assume 5% interest rate.

Solution:

526,25$])05.01[(000,20])1[( 5 =+=+×= niPF

27

Economic Efficiency

Case 2:

Description: Finding the initial amount (P) that would yield a future amount (F) at the end of a given period

Example: Jim has seen the 2012 model in a car magazine. Wishes to buy that model when it is released in 2012, at a price of $50,000 at that year.

How much should Jim put away now in order to be able to pay for the car in 2008?

XXZ, 2012 Edition

28

Economic Efficiency

29

Problem Definition: This is a Single Payment Present Worth Factor (SPPWF) problem

Cash Flow Diagram:

Computational Formula:

niFP

)1( +=

2007 2012

P F

Case 2 (continued):

SPCAF

Economic Efficiency


Question: Jim has seen the 2012 model in a car magazine. Wishes to buy that model when it is released in 2012, at a price of $50,000 at that year.

Solution:

176,39$)05.01(

000,50)1( 5 =+

=+

= niFP

30

Economic Efficiency

Case 3:

Description: Finding the amount of uniform annual payments (A) that would yield a certain future amount (F) at the end of a given period.

Example: Jim has seen the 2012 model in a car magazine. Wishes to buy that model when it is released in 2012, at a price of $50,000 at that year. Jim agrees to pay 5 yearly amounts until 2012, starting December 2008. How much should he pay every year?

31

Economic Efficiency

Case 3 (continued):

32

Problem Definition: This is a Uniform Series Sinking Fund DepositFactor (USSFDF) problem

Cash Flow Diagram:


2007 2012

A=?

F

1)1( −+×= ni

iFA

A=?A=?A=?A=?

SPCAF

Economic Efficiency


Question :Jim agrees to pay 5 yearly amounts, starting December 2007 until 2012. How much should he pay every year so he can end up paying full price ($50,000) in 2012?

Solution:

74.048,9$1)05.01(

05.0000,501)1( 5 =

−+×=

−+×= ni

iFA

33

Economic Efficiency

Case 4:

Description: Finding the final compounded amount (F) at the end of a given period due to uniform annual payments (A).

Example: Same as for Case 2 (Jim has seen the 2012 model in a car magazine, and wishes to purchase it when it is released in 2012).

Price of the car is not fixed. Rather, the Dealer and Jim agree that Jim will pay $9,000 every year starting December 2007 until 2012. How much will he end up paying for the car in 2012?

34

Economic Efficiency

Problem Definition: This is a Uniform Series Compounded AmountFactor (USCAF) problem

Cash Flow Diagram:


2007 2012

F = ?

iiAF

n 1)1( −+×=

A A A A A

Case 4 (continued):

SPCAF

Economic Efficiency


Question:Dealer and Jim agree that Jim will pay $9,000 every year starting December 2007 until 2012. How much will he end up paying for the car in 2012?

Solution:

68.730,49$05.0

1)05.01(000,91)1( 5

=−+

×=−+

×=iiAF

n

Economic Efficiency

Case 5:

Description: Finding the initial amount (P) that would yield specified uniform future amounts (A) over a given period.

Example: Jim takes the 2007 model now (in 2007). He has enough money to

pay for it, but rather decides to pay in annual installments of $5,000 over a 5-year period (starting in Dec 2007 till Dec 2012). How much should he set aside now so that he can make such annual payments?

37

Economic Efficiency

Case 5 (continued):

Problem Definition: This is a Uniform Series Present Worth Factor (USPWF) problem

Cash Flow Diagram:


2007 2012

P=?A A A AA

n

n

iiiAP

)1(1)1(

+−+

×=USPWF

Economic Efficiency


Question:Jim rather decides to pay in annual installments of $5,000 over a 5-

year period (starting in Dec 2007 till Dec 2012).

How much should he set aside now so that he can make such annualpayments?

Solution:

38.647,21$)05.01(05.01)05.01(000,5

)1(1)1(

5

5

=+

−+×=

+−+

×= n

n

iiiAP

39

Economic Efficiency

Case 6:

Description: Finding the amount uniform annual payments (A) over a given period, that would completely recover an initial amount (P) such as a loan

Example:

Jim receives a loan of $25,000 to pay for the 2007 model and take it right now. How much will he have to pay back to the bank every year (starting Dec 2008) until Dec 2012?

40

Economic Efficiency

Case 6 (continued):

Problem Definition: This is a Capital Recovery Factor (CRF) problem. (that is, the bank seeks to “recover its capital” from Jim).

Cash Flow Diagram:


2007 2012

PA=?

1)1()1(−+

+×= n

n

iiiPA

A=? A=? A=? A=?

CRF

41

Economic Efficiency


Question:Jim receives a loan of $25,000 to pay for the 2007 model and

take it right now. How much will he have to pay back to the bank every year (starting Dec 2008) until Dec 2012?

Solution:

37.774,5$1)05.01()05.01(05.0000,25

1)1()1(

5 =−+

+×=

−++

×=n

n

n

iiiPA

42

SummaryCash Flow Diagram

Computational Formula Factor Computation

P F=? 0 1 … … N

F = P × SPCAF The Single Payment Compound Amount Factor, SPCAF (i%, N).

NiSPCAF )1( +=

P=? F 0 1 … … N

P = F× SPPWF The Single Payment Present Worth Factor, SPPWF (i%, N).

NiSPPWF

)1(1+

=

A=? A=? A=? A=? 0 1 2 … N F

A = F × SFDF The Sinking Fund Deposit Factor, SFDF (i%, N).

1)1( −+= Ni

iSFDF

A A A A 0 1 2 … N F=?

F = A × USCAF The Uniform Series Compound Amount Factor, USCAF (i%, N).

iiUSCAF

N 1)1( −+=

P=? A A A A 0 1 2 … N

P = A× USPWF The Uniform Series Present Worth Factor, USPWF (i%, N).

N

N

iiiUSPWF

)1(1)1(

+×−+

=

P A=? A=? A=? A=? 0 1 2 … N

A = P × CRF The Capital Recovery Factor, CRF (i%, N).

1)1()1(−+

+×= N

N

iiiCRF

43

What if the Interest Rate, i is not fixed but is(i) Variable(ii) Compounded a fixed number of

times, m, during each year?

44

In that case, in place of i in the Interest Formula, we should use ie, the nominal interest rate

But how do we determine the value of ie?

11 −⎟⎠⎞

⎜⎝⎛ +=

mn

e mii

If:

in is the nominal annual interest rate, and

m is the number of compounding periods in a year

Then:

45

SUMMARY FOR FORMULA FOR:When the Interest Rate, i is not fixed but is (i) variable and (ii) compounded a fixed number of times, m, during each year?

Cash Flow Diagram

Computational Formula FaComp

P F=? 0 1 … … N


NiSPCAF )1( +=

P=? F 0 1 … … N


NiSPPWF

)1(1+

=

A=? A=? A=? A=? 0 1 2 … N F


1)1( −+= Ni

iSFDF

A A A A 0 1 2 … N F=?


iiUSCAF

N)1( −+=

P=? A A A A 0 1 2 … N


N

iiiUSPWF

)1()1(+×

−+=

P A=? A=? A=? A=? 0 1 2 … N


1)1()1(−+

+×= N

N

iiiCRF

Factor Computation

NeiSPCAF )1( +=

Nei

SPPWF)1(

1+

=

1)1( −+= N

eiiSFDF

ii

USCAFN

e 1)1( −+=

N

ee

Ne

iii

USPWF)1(1)1(

+×−+

=

1)1()1(−+

+×= N

e

Nee

iii

CRF

11 −⎟⎠⎞

⎜⎝⎛ +=

mn

e mii

Where:

Example:

(Case 1) Current car price is $20,000. Jim takes the car now (Dec 2007) and agrees with dealer to pay nothing until Dec 2012 when he pays everything at a go.

What price will he pay in 2012? Assume a nominal 5% interest rate compounded every quarter.

Solution:in is the nominal annual interest rate, andm is the number of compounding periods in a year

The final amount F can then be calculated as follows:21.635,25$])0509.01[(000,20])1[( 5 =+=+×= n

eiPF

Economic Efficiency

%09.50509.01405.0111

4

==−⎟⎠⎞

⎜⎝⎛ +=−⎟

⎠⎞

⎜⎝⎛ +=

mn

e mii

47

What if the Interest Rate, i is not fixed but is(i) Variable(ii) Compounded an infinite number of

times during each year?

48

Effective Interest Rate when: interest rate is variable and is being compounded an infinite number of time in a year

In that case, in place of i in the Interest Formula, we should use ie, the nominal interest rate

But how do we determine the value of ie?

1−= ie ei

If:

in is the nominal annual interest rate, and

m is the number of compounding periods in a year

Then:

Substituting ie for i in the original equations yield a new set of Interest Equations (next slide) 49

Interest Equations for Variable and Infinite CompoundingCash Flow Diagram

Computational Formula Factor Computation

P F=? 0 1 … … N


iNeSPCAF ×=

P=? F 0 1 … … N


iNeSPPWF ×=

1

A=? A=? A=? A=? 0 1 2 … N F


11−−

= × iN

i

eeSFDF

A A A A 0 1 2 … N F=?


11

−−

=×

i

iN

eeUSCAF

P=? A A A A 0 1 2 … N


11

−−

=×−

i

iN

eeUSPWF

P A=? A=? A=? A=? 0 1 2 … N


iN

i

eeCRF ×−−

−=

11

50

Some In-class questions …Five years from now, an airport authority intends to rehabilitate its runways at a cost of $1.5 million. Ten years from now, the runways will be replaced at a cost of $5.8 million.

Assuming an interest rate of 8%, find the total present worth ofthese costs.

51

SolutionFive years from now, an airport authority intends to rehabilitate its runways at a cost of $1.5 million. Ten years from now, the runways will be replaced at a cost of $5.8 million.

Assuming an interest rate of 8%, find the total present worth of these costs.

PW = 1.5M*SPPWF (8%, 5 years) + 5.8M*SPPWF (8%, 10 years) = 3.71M

PW = ? 1.5 M 5.8 M

5 5

52

Some In –class questions …A major corridor investment is expected to yield $50,000 per year in reduced crash costs, $200,000 per year in reduced vehicle operating costs, and $405,000 per year in travel time costs.

What is the combined present worth of these costs? Assume the interest rate is 5% and the analysis period is 20 years.

53

SolutionA major corridor investment is expected to yield $50,000 per year in reduced crash costs, $200,000 per year in reduced vehicle operating costs, and $405,000 per year in travel time costs.

What is the combined present worth of these costs? Assume the interest rate is 5% and the analysis period is 20 years.

PW = (50,000 + 20,000 + 405,000)* USPWF (5%, 20) = 5.92M

54

Criteria for Economic Efficiency Impact Evaluation

Present Worth of Costs (PWC)

Equivalent Uniform Annual Cost (EUAC)

Equivalent Uniform Annual Return (EUAR)

Net Present Value (NPV)

Internal Rate of Return (IRR)

Benefit-cost Ratio (BCR)

55

1. Present Worth of CostsIs used when benefits of all alternatives are equal, so cost is the only criterion to consider in choosing the best alternative

Alternatives have the same service life or analysis period

Converts all costs into an equivalent single cost assumed to occur at the beginning of the analysis period (time = zero).

56

Present Worth of Costs - Example

An airplane purchase is proposed.

For airplane type Ainitial cost = $50 million, average annual maintenance cost = $0.25 million, salvage value = $8 million.

For airplane type Binitial cost = $30 million, average annual maintenance cost = $0.75 million, salvage value = $2 million.

Both types have a useful life of 15 years. Which alternative should be selected? Assume 7% interest rate.

57

Solution

PWCA (in millions)

= 50 + 0.25*USPWF(7%, 15) – 8*SPPWF(7%, 15) = $49.38M

PWCB (in millions)

= 30 + 0.75*USPWF(7%, 15) – 2*SPPWF(7%, 15) = $36.11M

Alternative B has a lower PWC

Thus Alt B. is more desirable.58

2. Equivalent Uniform Annual Costs (EUAC)

Is used when benefits of all alternatives are equal, so cost is the only criterion to consider in choosing the best alternative

Alternatives have the different service lives or analysis periods

Converts all costs into an equivalent cost assumed to occur at each year of the analysis period.

59

ExampleBus transit services in MetroCity can be performed satisfactorily using any one of two alternative bus types, A or B.

Bus type A: initial cost = $100,000estimated life = 6 yrsannual maintenance & operating costs = $8,000salvage value = $20,000.

Bus type B: initial cost = $75,000estimated life = 5 yrsannual maintenance & operating costs = $8,000 for

the first 2yrs and $12,000 for the remaining 4 yrssalvage value = $10,000.

Find the equivalent annual cost of each alternative, and decide which option is more desirable. Assume a 6% interest rate.

60

SolutionEUACA (in thousands) = 100CRF(6%, 6) + 8USPWF(6%, 6) CRF(6%, 6) – 20SFDF(6%, 6) = $25.47

EUACB (in thousands) = 75CRF(6%, 5) + 8USPWF(6%, 2) CRF(6%, 6) + (12USPWF(6%, 4) SPPWF(6%, 2)CRF(6%, 6) – 40SFDF(6%, 6) = $22.57

Alternative B is more desirable.

61

Equivalent Uniform Annual Return

Used when costs are different and benefits are also different across alternatives

Combines all costs and benefits or returns associated into a single annual value of return (benefits less costs) over the analysis period.

This method can be used when the alternatives have different service lives.

62

ExampleTwo alternative designs proposed for renovating Water Port. Annual benefits Aare in terms of monetized savings in inventory delay, safety and security, and vessel operations.Alternative A: initial project cost = $200 million, life = 25 yrs,

salvage value = $22 million, annual maintenance/operating costs = $80 million, annual benefits = $75 million.

Alternative B: initial project cost = $175 million, life = 20 yrs, salvage value = $15 million, annual maintenance/operating costs =

$90 million, annual benefits = $55 million.

Find the equivalent uniform annual return of each alternative and identify the better alternative. Assume a 4% interest rate.

63

EUARA (in millions)

= 75 – 200CRF(4%,25) – 80 + 22SFDF(4%,25) = –$17.27M

EUARB (in millions)

= 55 – 175CRF(4%,25) – 90 + 15SFDF(4%,25) = –$45.84M

Alternative A is more desirable.

64

The NPV of an investment is the difference between the present worth of benefits and the present worth of costs.

NPV reflects the value of the project at the time of the base year of the analysis which may be considered the year of decision making.

NPV is often considered as the best economic efficiency indicator as it provides a magnitude of net benefits in monetary terms.

Among competing transportation projects or policies, the alternative with the highest NPV is considered the most “economically efficient.”

65

ExampleTwo alternative designs proposed for renovating Water Port. Annual benefits Aare in terms of monetized savings in inventory delay, safety and security, and vessel operations.Alternative A: initial project cost = $200 million, life = 25 yrs,

salvage value = $22 million, annual maintenance/operating costs = $80 million, annual benefits = $75 million.

Alternative B: initial project cost = $175 million, life = 20 yrs, salvage value = $15 annual maintenance/operating costs = $90

million, annual benefits = $55 million.

Find the NPV each alternative and identify the better alternative. Assume a 4% interest rate.

66

NPVA (in millions)

= 75 *USPWF(4%, 25) – 200 – 80*USPWF(4%, 25) + 22*SPPWF(4%,25) = – $269.86M

NPVB (in millions)

= 55* USPWF(4%, 20) – 175 – 90*USPWF(4%, 20) + 15*SPPWF(4%,20) = – $716.15M

Alternative A is more desirable.

67

The Internal Rate of Return (IRR)IRR found by equating PWbenefits to PWcosts, or by equating EUACbenefits to EUACcosts.

Minimum attractive rate of return (MARR)

is the lowest rate of return that investors will accept before they invest considering the likely investment risks or the opportunity to invest elsewhere for possibly greater returns.

Then the IRR is compared to the Minimum Attractive Rate of Return (MARR).

If the IRR > MARR, then the investment is considered worthwhile.If the IRR < MARR, then investment is considered NOT worthwhile.

68

ExampleAn urban rail transit agency is considering the purchase of a

new $30,000 ticketing system that will reduce travel time.

The estimated life of the system is 10 years at which time the value of the system will be $15,000.

The expected travel time savings per year is $5,000 per year, and the average annual maintenance and operating cost is $2,000.

Is the project economically more desirable than the do-nothing alternative? The minimum attractive rate of return is 5%.

SolutionEquating the net cash flow on both sides yields:

5,000USPWF(i%,10) + 15,000SPPWF(i%,10) = 30,000 + 2000USPWF(i%,10)

Solving this equation by trial and error or using Microsoft Excel yields: i = 6.25% This value exceeds the MARR value of 5%.

So it is economically more efficient to undertake the project than to do nothing.

Benefit Cost Ratio

The benefit-cost ratio (BCR) is a ratio of benefits to costs.

That is:

NPVBenefits/NPBCosts, OR

EUABenefits/EUACosts

An investment with a BCR exceeding 1 is considered to be economically feasible

The alternative (investment) with the highest BCR value is considered the best.

72

Example

73

Two alternative designs proposed for renovating Water Port.

Annual benefits Aare in terms of monetized savings in inventory delay, safety and security, and vessel operations.

Alternative A: initial project cost = $200 million, life = 25 yrs, salvage value = $22 million, annual maintenance/operating costs = $80 million, annual benefits = $75 million.

Alternative B: initial project cost = $175 million, life = 20 yrs,

salvage value = $15 annual maintenance/operating costs = $90 million, annual benefits = $55 million.

Find the equivalent uniform annual return of each alternative and identify the better alternative. Assume a 4% interest rate.

SolutionTaking the ratio of all benefits to all costs yields:

For Alternative A:

For Alternative B:

BCRA > BCRB, hence, A is a better alternative.

75 (4%,25) 22 (4%,25) 0.8139200 80 (4%,25)

AA

A

PWB USPWF SPPWFBCRPWC USPWF

× + ×= = =

+ ×

55 (4%,25) 15 (4%,25) 0.5470175 90 (4%,25)

BB

B

PWB USPWF SPPWFBCRPWC USPWF

× + ×= = =

+ ×

74

What’s your opinion …?

NPV vs. BCR

Which is better?

Special Interest-Equations

- Comparing Alternatives with Different Service Lives

- Present Worth of Continuously Compounded Payments with Continuously Compounded Interest

Special Interest Equations:Comparing Alternatives with Different Service Lives

Approaches

1. For each alternative, convert all costs and benefits into EUAR, OR

2. For alternative, find NPW over a service life that is a LCD (lowest common denominator), OR

3. Present Worth of Periodic Payments to Perpetuity

Approach 2 for Evaluating Alternatives with Different Service Lives

For service lives n1 and n2, determine the lowest common denominator, n* and repeat the entire cost stream over n*, for both alternatives.

n1 = 3 years n2 = 4 yearsAlternative 1 Alternative 2

Assume total replacement (reconstruction) of the facility after the end of its service life

n* = 12 yearsAlternative 1

n* = 12 yearsAlternative 2

With the same analysis period of 12 years, we can now proceed to use NPV to evaluate the two alternatives


Present Worth of Periodic Payments in Perpetuity

For each alternative i: For one service life only, convert all costs and benefits over service life N, to a single compounded amount, R at end of service life find the present worth of R payments every N years over an analysis period of infinity

Assume total replacement (reconstruction) of the facility after the end of its service life, forever

N yrs N yrs N yrs N yrs

P’ R R R RR

RP’ P’


Present Worth of Periodic Payments in Perpetuity

For each alternative i: For one service life only, convert all costs and benefits over service life N, to a single present worth, R and find the present worth of R payments every ni years over an analysis period of infinity

...)1()1()1(

' 32, ++

++

++

+=∞ NNNR iR

iR

iRPPW

⎥⎦

⎤⎢⎣

⎡+

++

++

++= ...

)1(1

)1(1

)1(1' 32 NNN iii

RP

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

+−

+= 1

)1(11

1'Ni

RP 1)1('

−++= Ni

RP

Assume total replacement (reconstruction) of the facility after the end of its service life, forever

Example:A railway bridge is slated for reconstruction.

Estimated service life of the structure is 60 years.

Reconstruction cost is $600,000.

During its replacement cycle, the bridge will require two rehabilitations, each with a cost of $200,000, at the 20th and 40th year

Average annual cost of maintenance is $5,000.

At the end of the replacement cycle, the bridge will again be reconstructed and the entire cycle is assumed to repeat in perpetuity.

Find the present worth of all costs to perpetuity? Assume 5% interest rate. Assume P’ (non-recurring cost = 0)

Solution:

C = $600,000 $200,000 $200,000

0 20 yr 40 yr 60 yr

$5000/yr

R

R = compounded life-cycle cost= 600,000SPCAF(5%, 60) + 200,000SPCAF(5%, 40) + 200,000SPCAF(5%, 20) + 5,000USCAF(5%, 60) = $14,914,087

R R

0 60 yrs 120 yrs∞

1)1( −+ NiR

( )596,843$

105.01087,914,1460 =−+

=PWR∞ =

60 years


Why is the perpetuity issue particularly relevant to transportation

systems evaluation?

Why is perpetuity particularly relevant to transportation systems evaluation?

Because:

Transportation facilities are needed by society forever

Transportation facilities do not last foreverHence need replacement after some decades

Special Interest Equations: Present Worth of Continuously Compounded Payments with Continuously Compounded Interest

Consider a general case where an initial amount, R0, grows exponentially at a rate of r expressed as a percentage per year.

Bringing all future streams to the present gives:

R0: Present worth of R0 = R0 = R

R1: Present worth of R1 = (R × er)/ei

R2: Present worth of R2 = (R × e2r)/e2i

Rn-1: Present worth of R-1n = (R × e(n-1)r)/e(n-1)i

R0 R1 R2

R3

1st year 2nd year 3rd year

Rn-1

nth year 1 3 n–1 n 2 0

⎥⎦

⎤⎢⎣

⎡−−

×=⎥⎦

⎤⎢⎣

⎡++++×=

−

−

−

)(1...1

)(

)(

)(

2

2

, ireR

ee

ee

eeRPW

irn

iin

rin

i

r

i

r

CCICCP r < i

Example

160,719$)10.003.0(1000,100

)(1 )10.003.0(10)(

, =⎥⎦

⎤⎢⎣

⎡−

−×=⎥

⎦

⎤⎢⎣

⎡−−

×=−− e

ireRPW

irn

CCICCP

The average annual cost of operating the physical infrastructure of a small airport facility is currently $100,000.

Due to growth in air traffic, the annual costs are compounded continuously at 3% per annum.

What is the present worth of the operating costs over a period of 10 years? Assume a 10% interest rate that is compounded continuously

Solution


Present Worth of

Continuously Compounded Payments with

Continuously Compounded Interest

Is this really an issue?

Special Interest Equations: Present Worth of Continuously Compounded Payments with Continuously Compounded Interest

Is this really an issue?Annual costs not really uniform but typically grow with time

(i) Systems get olderCurrent year’s condition depend on previous years conditionCosts more to preserveCosts are compounded as the years go by

(ii) Traffic grows with timeAll costs link to traffic use (annual operating costs are not constant but increase (compounded over time)


89


90

1. Identify the Characteristics of the Transportation Project

2. Identify the Purpose of the Economic Analysis

3. Select and Describe the Base CaseandtheOther Alternatives4. Define and Describe

the Study Area 5. Select the Appropriate

Analysis Period for the Study

6. Identify the Appropriate Impact Types for Economic

8. Determine Characteristics of the Transportation System under theVariousAlternativeScenarios

7. Select the Appropriate Economic Analysis Criterion

9. Apply Data to Estimate the Economic Efficiency Impacts

Software Packages for Economic Efficiency Analysis

Surface Transportation Efficiency Analysis Model (STEAM)

MicroBenCost

Highway Development and Management (HDM) Standards Model — World Bank

Highway Economic Requirements (HERS)

Indiana DOT’s Major Corridor Investment-Benefit Analysis System (MCIBAS)

California DOT’S Cal B/C System

Software Packages for Economic Efficiency Analysis

MicroBenCost Demonstration

What is Life-cycle Cost Analysis (LCCA)?Assessing the impacts of a transportation system over its service life. LCCA is relevant for new facilities.

Service life:Construction to Demolition/SalvageOften in years

Involves both costs and benefits ($)

Includes cost of operations, routine maintenance, periodic maintenance, etc.

Helps us decide over the service life:what activities to do When to do those activities

What is Remaining Service Life (RSL) Analysis?

Sometimes, facility already exists, has some more years left to serve. RSL is relevant for existing facilities.

And we need to decide over its remaining service life:what activities to do when to do those activities

Thus, we assess the cost and benefits of transportation system over its remaining service life

Most infrastructure are existing. So RSL analysis very relevant.

94

Reading Assignment

Is it really “fair” to evaluate projects on basis of only Economic Efficiency? Some thoughts (Pages 212-213 of the Text)

95

evaluating the economic efficiency of transportation ...srg/book/files/pdf/8....

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